Object a has a mass of 20 g
1 answer:
You might be interested in
Answer:
The frequency of the oscillation is 2.45 Hz.
Explanation:
Given;
mass of the spring, m = 0.5 kg
total mechanical energy of the spring, E = 12 J
Determine the spring constant, k as follows;
E = ¹/₂kA²
kA² = 2E
k = (2E) / (A²)
k = (2 x 12) / (0.45²)
k = 118.519 N/m
Determine the angular frequency, ω;
Determine the frequency of the oscillation;
ω = 2πf
f = (ω) / (2π)
f = (15.396) / (2π)
f = 2.45 Hz
Therefore, the frequency of the oscillation is 2.45 Hz.
Answer:
a)
b)
c)
d)
e)
Explanation:
1) Important concepts
Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".
2) Part a
The equation that describes the simple armonic motion is given by (1)
And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.
For the velocity:
(2)
For the acceleration
(3)
As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:
Since we know the amplitude A=0.002m we can solve for like this:
And we with this value we can find the period with the following formula
3) Part b
From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:
4) Part c
In order to find the total mechanical energy of the oscillator we can use this formula:
5) Part d
When we want to find the force from the 2nd Law of Newton we know that F=ma.
At the maximum displacement we know that X=A, and in order to that happens , and we also know that the maximum acceleration is given by::
So then we have that:
And since we have everything we can find the force
6) Part e
When the mass it's at the half of it's maximum displacement the term and on this case the acceleration would be given by;
And the force would be given by:
And replacing we have:
Answer:
The answer to the question
The steady state response is u₂(t) = -cos(3t + π/4)
of the form R·cos(ωt−δ) with R = , ω = 3 and δ = -π/4
Explanation:
To solve the question we note that the equation of motion is given by
m·u'' + γ·u' + k·u = F(t) where
m = mass = 2.00 kg
γ = Damping coefficient = 1
k = Spring constant = 3 N·m
F(t) = externally applied force = 27·cos(3·t)−18·sin(3·t)
Therefore we have 2·u'' + u' + 3·u = 27·cos(3·t)−18·sin(3·t)
The homogeneous equation 2·u'' + u' + 3·u is first solved as follows
2·u'' + u' + 3·u = 0 where putting the characteristic equation as
2·X² + X + 3 = 0 we have the solution given by =
This gives the general solution of the homogeneous equation as
u₁(t) =
For a particular equation of the form 2·u''+u'+3·u = 27·cos(3·t)−18·sin(3·t) which is in the form u₂(t) = A·cos(3·t) + B·sin(3·t)
Then u₂'(t) = -3·A·sin(3·t) + 3·B·cos(3·t) also u₂''(t) = -9·A·cos(3·t) - 9·B·sin(3·t) from which 2·u₂''(t)+u₂'(t)+3·u₂(t) = (3·B-15·A)·cos(3·t) + (-3·A-15·B)·sin(3·t). Comparing with the equation 27·cos(3·t)−18·sin(3·t) we have
3·B-15·A = 27
3·A +15·B = 18
Solving the above linear system of equations we have
A = -1.5, B = 1.5 and u₂(t) = A·cos(3·t) + B·sin(3·t) becomes 1.5·sin(3·t) - 1.5·cos(3·t)
u₂(t) = 1.5·(sin(3·t) - cos(3·t) = ·cos(3·t + π/4)
The general solution is then u(t) = u₁(t) + u₂(t)
however since u₁(t) = ⇒ 0 as t → ∞ the steady state response = u₂(t) =
·cos(3·t + π/4) which is of the form R·cos(ωt−δ) where
R =
ω = 3 and
δ = -π/4